不过可能是个人应用的只是简单的一个对数据的加密,所以感觉目前和MD5差距并不是很大.
1.首先要找到一个合适的加密工具类(网上一搜一大堆),我的放在最下面了
2.我目前的应用步骤
我这边先使用时候,我会根据id+key(这两个都是生成的Longl类型的串)储在数据库中
将两者拼接在一起的字符串使用工具类加密,返还给接收数据的页面
查询数据的时候,调用的地方会给我回传id,和一个密匙
我首先先通过id,查询出数据库真实的数据,之前数据在存储数据库中会存储一个key,
拿出这个ke与id按照之前的加密方式在加密一次,之后与前端页面传来的密匙进行比较,
当两者一致时候,我把正确的数据回传给调用者,不一致则提示密匙错误
因为是第一次使用sha256,所以只是简单的应用,可能与每个人的应用不一样,也可能我的使用方式错了,请您指出来,我好及时改正
//下面程序由520huiqin编写，已在VC++ 6.0下编译通过#include <iostream.h>
#include <math.h>
#include <stdio.h>
typedef int Elemtype
Elemtype p,q,e
Elemtype fn
Elemtype m,c
int flag = 0
typedef void (*Msghandler) (void)
struct MsgMap {
char ch
Msghandler handler
}
/* 公钥 */
struct PU {
Elemtype e
Elemtype n
} pu
/* 私钥 */
struct PR {
Elemtype d
Elemtype n
} pr
/* 判定一个数是否为素数 */
bool test_prime(Elemtype m) {
if (m <= 1) {
return false
}
else if (m == 2) {
return true
}
else {
for(int i=2i<=sqrt(m)i++) {
if((m % i) == 0) {
return false
break
}
}
return true
}
}
/* 将十进制数据转化为二进制数组 */
void switch_to_bit(Elemtype b, Elemtype bin[32]) {
int n = 0
while( b >0) {
bin[n] = b % 2
n++
b /= 2
}
}
/* 候选菜单，主界面 */
void Init() {
cout<<"*********************************************"<<endl
cout<<"*** Welcome to use RSA encoder ***"<<endl
cout<<"*** a.about ***"<<endl
cout<<"*** e.encrypt ***"<<endl
cout<<"*** d.decrypt ***"<<endl
cout<<"*** s.setkey***"<<endl
cout<<"*** q.quit ***"<<endl
cout<<"**********************************by*Terry***"<<endl
cout<<"press a key:"<<endl
}
/* 将两个数排序，大的在前面*/
void order(Elemtype &in1, Elemtype &in2) {
Elemtype a = ( in1 >in2 ? in1 : in2)
Elemtype b = ( in1 <in2 ? in1 : in2)
in1 = a
in2 = b
}
/* 求最大公约数 */
Elemtype gcd(Elemtype a, Elemtype b) {
order(a,b)
int r
if(b == 0) {
return a
}
else {
while(true) {
r = a % b
a = b
b = r
if (b == 0) {
return a
break
}
}
}
}
/* 用扩展的欧几里得算法求乘法逆元 */
Elemtype extend_euclid(Elemtype m, Elemtype bin) {
order(m,bin)
Elemtype a[3],b[3],t[3]
a[0] = 1, a[1] = 0, a[2] = m
b[0] = 0, b[1] = 1, b[2] = bin
if (b[2] == 0) {
return a[2] = gcd(m, bin)
}
if (b[2] ==1) {
return b[2] = gcd(m, bin)
}
while(true) {
if (b[2] ==1) {
return b[1]
break
}
int q = a[2] / b[2]
for(int i=0i<3i++) {
t[i] = a[i] - q * b[i]
a[i] = b[i]
b[i] = t[i]
}
}
}
/* 快速模幂算法 */
Elemtype modular_multiplication(Elemtype a, Elemtype b, Elemtype n) {
Elemtype f = 1
Elemtype bin[32]
switch_to_bit(b,bin)
for(int i=31i>=0i--) {
f = (f * f) % n
if(bin[i] == 1) {
f = (f * a) % n
}
}
return f
}
/* 产生密钥 */
void produce_key() {
cout<<"input two primes p and q:"
cin>>p>>q
while (!(test_prime(p)&&test_prime(q))){
cout<<"wrong input,please make sure two number are both primes!"<<endl
cout<<"input two primes p and q:"
cin>>p>>q
}
pr.n = p * q
pu.n = p * q
fn = (p - 1) * (q - 1)
cout<<"fn = "<<fn<<endl
cout<<"input e :"
cin>>e
while((gcd(fn,e)!=1)) {
cout<<"e is error,try again!"
cout<<"input e :"
cin>>e
}
pr.d = (extend_euclid(fn,e) + fn) % fn
pu.e = e
flag = 1
cout<<"PR.d: "<<pr.d<<" PR.n: "<<pr.n<<endl
cout<<"PU.e: "<<pu.e<<" PU.n: "<<pu.n<<endl
}
/* 加密 */
void encrypt() {
if(flag == 0) {
cout<<"setkey first:"<<endl
produce_key()
}
cout<<"input m:"
cin>>m
c = modular_multiplication(m,pu.e,pu.n)
cout<<"c is:"<<c<<endl
}
/* 解密 */
void decrypt() {
if(flag == 0) {
cout<<"setkey first:"<<endl
produce_key()
}
cout<<"input c:"
cin>>c
m = modular_multiplication(c,pr.d,pr.n)
cout<<"m is:"<<m<<endl
}
/* 版权信息 */
void about() {
cout<<"*********************************************"<<endl
cout<<"*** by Terry***"<<endl
cout<<"*** copyright 2010,All rights reserved by ***"<<endl
cout<<"*** Terry,technology supported by weizuo !***"<<endl
cout<<"*** If you have any question, please mail ***"<<endl
cout<<"*** to 18679376@qq.com !***"<<endl
cout<<"*** Computer of science and engineering ***"<<endl
cout<<"*** XiDian University2010-4-29***"<<endl
cout<<"*********************************************"<<endl
cout<<endl<<endl
Init()
}
/* 消息映射 */
MsgMap Messagemap[] = {
{'a',about},
{'s',produce_key},
{'d',decrypt},
{'e',encrypt},
{'q',NULL}
}
/* 主函数，提供循环 */
void main() {
Init()
char d
while((d = getchar())!='q') {
int i = 0
while(Messagemap[i].ch) {
if(Messagemap[i].ch == d) {
Messagemap[i].handler()
break
}
i++
}
}
}
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